What is the integral of (sin^2(1/t))/(t^2) ?

The integral of (sin^2(1/t))/(t^2) is 1/2 (-1/t+1/2 sin(2/t))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sin^2(1/t))/(t^2)

\int \frac{sin ^{2} ( \frac{1}{t} )}{t ^{2}} d t

\frac{1}{2} (- \frac{1}{t} + \frac{1}{2} sin (\frac{2}{t})) + C

Solution steps

\int \frac{sin ^{2} ( \frac{1}{t} )}{t ^{2}} d t

Rewrite using trig identities

= \int \frac{1 - cos ( \frac{2}{t} )}{2 t ^{2}} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1 - cos ( \frac{2}{t} )}{t ^{2}} d t

Expand

\frac{1 - cos ( \frac{2}{t} )}{t ^{2}} : \frac{1}{t ^{2}} - \frac{cos ( \frac{2}{t} )}{t ^{2}}

= \frac{1}{2} \cdot \int \frac{1}{t ^{2}} - \frac{cos ( \frac{2}{t} )}{t ^{2}} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{1}{t ^{2}} d t - \int \frac{cos ( \frac{2}{t} )}{t ^{2}} d t)

\int \frac{1}{t ^{2}} d t = - \frac{1}{t}

\int \frac{cos ( \frac{2}{t} )}{t ^{2}} d t = - \frac{1}{2} sin (\frac{2}{t})

= \frac{1}{2} (- \frac{1}{t} - (- \frac{1}{2} sin (\frac{2}{t})))

Simplify

= \frac{1}{2} (- \frac{1}{t} + \frac{1}{2} sin (\frac{2}{t}))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{t} + \frac{1}{2} sin (\frac{2}{t})) + C

n = 0 \sum \infty (- 5 x)^{n}

\frac{d}{d t} ((\frac{t ^{3}}{4} - 2) + (\frac{4}{t} - 3) + cos (t - 2))

\int \frac{- 4 x ^{3}}{2 x ^{3} + x ^{2} - 2 x - 1} d x

x \to 1 lim (5 x^{4} - 5 x^{2} + 10)

f^{^{'}} (x) = f (x)^{4} \cdot (4 x^{2} + x)

What is the integral of (sin^2(1/t))/(t^2) ?
The integral of (sin^2(1/t))/(t^2) is 1/2 (-1/t+1/2 sin(2/t))+C